福建省 、永安一中、漳平一中三校2022-2023学年高三数学上学期12月联考试卷（Word版附答案）

“德化一中、永安一中、漳平一中”三校协作2022—2023学年第一学期联考高三数学试题（考试时间：120分钟总分：150分）第Ⅰ卷（选择题，共60分）一、单选题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的．1．若复数z满足(z1)i1i，则|z|（）A．iB．1iC．2D．12．已知集合Ax0log2x1，Bxxa.若ABA，则实数a的取值范围是（）A．aa1B．aa1C．aa2D．aa223．已知为实数，则使命题“x0(0,2)，2x0x010”是假命题的一个充分不必要条件是（）A．22B．13C．0D．44．若等边三角形ABC的边长为1，点M满足CMCB2CA，则MAMB（）A．3B．2C．23D．35．《九章算术》中将底面为直角三角形且侧棱垂直于底面的三棱柱称为“堑堵”；底面为矩形，一条侧棱垂直于底面的四棱锥称为“阳马”；四个面均为直角三角形的四面体称为“鳖臑”．如图，在堑堵ABCABC中，ACBC，且AAAB2，则下列说法错误的是（）1111A．四棱锥BAACC为“阳马”11B．四面体ACCB为“鳖臑”112C．四棱锥BAACC体积的最大值为113D．过A点分别作AEAB于点E，AFAC于点F，则EFAB11126．已知函数f(x)23cosxcosx2sinx，若f(x)在区间m,上单调递减，则实数m的最小值24是（）A．B．C．D．126126237．抛物线x4y的焦点为F，过点F作斜率为的直线l与抛物线在y轴右侧的部分相交于点A，3过点A作抛物线准线的垂线，垂足为H，则AHF的面积是（）A．43B．4C．33D．88．艾萨克·牛顿英国皇家学会会长，英国著名物理学家，同时在数学上也有许多杰出贡献，牛顿用“作切线”的方法求函数f(x)零点时给出一个数列x：满足n三校联考高三数学试题第1页共4页 fxnxx2n1n，我们把该数列称为牛顿数列．如果函数f(x)axbxc（a0）有两个零点1，fxnx2n2，数列xn为牛顿数列．设anln，已知a11，xn2，an的前n项和为Sn，则S20221x1n等于（）20232022A．2022B．2023C．2D．2二、多选题：本题共4小题，每小题5分，共20分．在每小题给出的选项中，有多项符合题目要求．全部选对的得5分，有选错的得0分，部分选对的得2分．9．已知函数f(x)对xR都有f(x)f(x4)f(2)，若函数yf(x3)的图象关于直线x3对称，且对x1,x2[0,2]，当x1x2时，都有（x2x1)f(x2)f(x1)0，则下列结论正确的是（）A．f(2)0B．f(x)是偶函数C．f(x)是周期为4的周期函数D．f(3)f(4)10．下列结论中，正确的结论有（）aamA．若ab0，m0，则B．如果0x1，那么x43x取得最大值时x的值为1bbmC．已知ab1，则alnbblnaD．若xxy2y，x0，y0，则x2y的最小值是811．如图，平面四边形ABCD中，△BCD是等边三角形，ABBD，且ABBD2，M是AD的中点．沿BD将△BCD翻折，折成三棱锥CABD，在翻折过程中，下列结论正确的是（）A．棱CD上总会有一点N，使得MN//平面ABCB．存在某个位置，使得CM与BD所成角为锐角C．CMB一定是二面角CADB的平面角28D．当平面ABD平面BDC时，三棱锥CABD的外接球的表面积是32212．在平面直角坐标系xoy中，已知点M(2,1)，N(2,1)，动点P满足PMPNa(aR)，记点P的轨迹为曲线C，则（）aA.存在实数a，使得曲线C上所有的点到点(1,)的距离大于24B.存在实数a，使得曲线C上有两点到点(5,0)与(5,0)的距离之和为6C.存在实数a，使得曲线C上有两点到点(5,0)与(5,0)的距离之差为2D.存在实数a，使得曲线C上有两点到点(a,0)的距离与到直线xa的距离相等三校联考高三数学试题第2页共4页 第Ⅱ卷（选择题，共90分）三、填空题：本题共4小题，每小题5分，共20分．2213．若点P(1,1)为圆xy6x0的弦MN的中点，则弦MN所在直线方程为.14．若等差数列{a}的公差为2，a是a与a的等比中项，则该数列的前n项和S取得最大值时，nn526n的值为.2x2x215．已知函数f(x)，若方程f(x)k有两个不相等的实数根，则实数k的取值范围xe是.2222xyab16．已知双曲线C:1(ab0)的左、右焦点分别为F，F，P是C上的一点，且Q(,0)2212ab2满足FPQ，FPQ，则双曲线C的离心率为.1262四、解答题：本题共6小题，共70分．解答应写出文字说明，证明过程或演算步骤．17．（本题满分10分）已知ABC的内角A,B,C的对边分别为a,b,c，且向量m(2ac,b)与向量n(cosC,cosB)共线.（Ⅰ）求B；33（Ⅱ）若b3，ABC的面积为，判断ABC的形状，并说明理由.418．（本题满分12分）数列{a}的前n项和S满足S2an.nnnn（Ⅰ）求数列{a}的通项公式；n（Ⅱ）若数列{b}为等差数列，且ba，ba，求数列{ab}的前n项T.n3273nnn19．（本题满分12分）如图，在四棱锥PABCD中，底面ABCD为直角梯形，AD//BC，ADC90，平面PAD底面ABCD，1Q为AD的中点，M是棱PC上的点，PAPD2，BCAD1，CD3．2（Ⅰ）求证：平面MQB平面PAD；（Ⅱ）若二面角MBQC的大小为60，求QM的长．三校联考高三数学试题第3页共4页 20．（本题满分12分）疫情期间，为保障学生安全，要对学校进行消毒处理．校园内某区域由矩形OABC与扇形OCD组成，ππOA2m，AB23m，COD．消毒装备的喷射角EOF，阴影部分为可消毒范围，要求点E33在弧CD上，点F在线段AB上，设FOC，可消毒范围的面积为S．（Ⅰ）求消毒面积S关于的关系式，并求出tan的范围；（Ⅱ）当消毒面积S最大时，求tan的值．21．（本题满分12分）22xy已知椭圆C:221(ab0)的左、右焦点分别为F1,F2，其焦距为23，点E在椭圆C上，EF1EF2，abb直线EF的斜率为（c为半焦距）．1c（Ⅰ）求椭圆C的方程；22（Ⅱ）设圆O:xy2的切线l交椭圆C于A,B两点（O为坐标原点），求证：OAOB；2（Ⅲ）在（Ⅱ）条件下，求|OA||OB|的最大值．22．（本题满分12分）12已知函数fxalnxxa1xa0．2（Ⅰ）讨论函数fx的单调性；（Ⅱ）设函数gx3axfx有两个极值点x1，x2x1x2，证明：gx1gx210lna．三校联考高三数学试题第4页共4页 “德化一中、永安一中、漳平一中”三校协作2022—2023学年第一学期联考高三数学试题参考答案一、选择题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的．1．D2．B3．C4．D5．C6．B7．A8．D二、选择题：本题共4小题，每小题5分，共20分．在每小题给出的选项中，有多项符合题目要求。全部选对的得5分，有选错的得0分，部分选对的得2分．9．ABC10．AD11．AD12．BD三、填空题：本题共4小题，每小题5分，共20分．264313．y2x114．615．,0{}16．225ee四、解答题：本题共6小题，共70分．解答应写出文字说明，证明过程或演算步骤．→→17.解：（Ⅰ）∵向量�=（2a﹣c，b）与向量�=（cosC，cosB）共线，∴（2a﹣c）cosB﹣bcosC＝0，········································································2分即(2sinAsinC)cosBsinBcosC，∴2sinAcosB＝sin（B+C）＝sinA，�∵sinA≠0，∴cosB=，···············································································4分��∵B∈（0，π），∴B=；··············································································5分�1333（Ⅱ）由已知SacsinBac，所以ac3································6分ABC2442222由余弦定理得ac2accos3，所以ac6······································7分3解得ac3，························································································9分�又因为B=，所以ABC为正三角形．·························································10分�三校联考高三数学试题参考答案第1页共7页 18.解：（Ⅰ）当�=�时，��=���−�，所以��=�，··········································1分因为��=���−�①，所以当�≥�时，��−�=���−�−�−�②，······················2分①-②得��=���−���−�−�，所以��=���−�+�，········································3分��+����−�+�+����−�+�所以===�，·····························································5分��−�+���−�+���−�+�所以��+�是首项为2，公比为2的等比数列，所以��+�=�⋅��−�，所以��=��−�；·······················································6分（Ⅱ）由（Ⅰ）知，��=�，��=�，所以��=��=�，��=��=�，···················7分设��的公差为�，则��=��+�−�⋅�，所以�=�，····································8分所以��=��+�−�⋅�=�，·····································································9分所以��=���−�=�⋅��−�，��设数列�⋅��的前�项和为�，�所以�=�×�+�×��+�×��+⋯+�⋅��③，���=�×��+�×��+�×��+⋯+�⋅��+�④，�③-④得����+���−���+��+�−��=�+�+�+⋯+�−�⋅�=−�⋅�=�−�⋅�−�，�−�所以��=�−�⋅��+�+�，······································································10分��+�又因为数列�的前�项和等于�+�+�+⋯+�=，································11分��+���+�所以����的前�项和为Tn=�−�⋅�−+�.····································12分�119.（Ⅰ）证明：∵AD//BC，BC=AD，Q为AD的中点，2∴四边形BCDQ为平行四边形，∴CD//BQ······················································2分∵∠ADC=90°，∴∠AQB=90°，即QB⊥AD，又∵平面PAD⊥平面ABCD，平面PAD∩平面ABCD=AD，∴BQ⊥平面PAD，∵BQ平面MQB，∴平面MQB⊥平面PAD；···············································5分三校联考高三数学试题参考答案第2页共7页 （Ⅱ）解：∵PA=PD，Q为AD的中点，∴PQ⊥AD，∵平面PAD⊥平面ABCD，且平面PAD∩平面ABCD=AD，∴PQ⊥平面ABCD，···6分如图，以Q为原点建立空间直角坐标系，则Q(0,0,0)，A(1,0,0)，P(0,0,3)，B(0,3,0)，C(1,3,0)，设PMPC，且01，得M(,3,33)，所以QM(,3,33)，QB(0,3,0)，···········································7分设平面MBQ法向量为m(x,y,z)，mQM0x3y(33)z0由，得，mQB03y01令x3，则m(3,0,)，··································································8分由题意知平面BQC的一个法向量为n(0,0,1)，··············································9分1|||nm|1∵二面角M-BQ-C为60°，∴cos60，nm1223()1解得，·····························································································11分21337∴QM(,,)，∴|QM|.··························································12分2222π20.解：（Ⅰ）由题意可知COE，31π2则扇形COE的面积为S1232π6，·····································3分232OC//AB，则AFOFOC，且AF，············································4分tan122∴梯形OCBF的面积为S243243，··································6分2tantan2∴SS1S22π436，tanOA3πtanABO，∴ABO，AB36三校联考高三数学试题参考答案第3页共7页 πππ又EOF，所以，3633∴tan,3；·····················································································8分323（Ⅱ）设y6，tan,3，tan326sin2223sin13sin113y6，且sin,，············9分22222sinsinsin3记为锐角，且sin，3π2当时，y0，此时函数y6单调递减，6tanπ2当时，y0，此时函数y6单调递增，································11分3tan36sin2∴当sin时，cos，y取最小值，S取最大值，此时tan.33cos2···············································································································12分21.解：（Ⅰ）连接EF1，EF2，由题意知FEF,c3，······························1分122bbc设EFF,tan,sin,cos，12caacbEF23,EF23,EFEF2a，1212aa623b22即2a,又ab3,解得a6,b3，aa22xy椭圆C的方程为1··········································································4分63（Ⅱ）（i）当切线与坐标轴垂直时,交点坐标为(2,2),AOB90,OAOB，·················································································································5分(ii)当切线与坐标轴不垂直时，设切线为ykxm(k0),A(x1,y1),B(x2,y2),三校联考高三数学试题参考答案第4页共7页 m由圆心到直线距离为d2,m222k2,·········································6分2k1222联立切线方程与椭圆方程消去y整理得(2k1)x4kmx2m60,24km2m6∴xx,xx，······························································7分1221222k12k122223m66kxxyy(k1)xxkm(xx)m0，1212121222k1OAOB，·····························································································9分综上所述，OAOB；（Ⅲ）当切线与坐标轴直时，OAOB4，当切线与坐标轴不垂直时,由（Ⅱ）知OAOB,OAOB2AB,················10分2222(1k)(8k2)AB1kxx1222k1，22令tk,则AB2231，······················································11分4t4t2当且仅当时k等号成立，OAOB32，2综上所述，OAOB的最大值为32.···························································12分22.解：（Ⅰ）由f(x)定义域为x0,，且2axa1xax1xafxxa1，·······································1分xxx令fx0得，x1或xa，①当0a1时，x0,a，f(x)0，fx单调递增，xa,1，f(x)0，fx单调递减，x1,，f(x)0，fx单调递增，···········································2分三校联考高三数学试题参考答案第5页共7页 ②当a1时，f(x)0，fx在0,单调递增，···········································3分③当a1时，x0,1，f(x)0，fx单调递增，x1,a，f(x)0，fx单调递减，xa,，f(x)0，fx单调递增，·············································4分综上，当0a1时，fx在区间0,a，1,上单调递增，fx在区间a,1上单调递减；当a1时，fx在区间0,上单调递增；当a1时，fx在区间0,1，a,上单调递增，fx在区间1,a单调递减；2212a4xaxx4xa（Ⅱ）由已知，gx4xalnxx，则gx4x，2xxx函数gx有两个极值点x,xxx，即x24xa0在0,上有两个不等实根，1212·················································································································5分h0a02令hxx4xa，只需，故0a4，····································6分h2a40又x1x24，x1x2a，················································································7分1212所以gx1gx24x1alnx1x14x2alnx2x2221224x1x2alnx1lnx2x1x2aalna8，···············8分2要证gx1gx210lna，即证aalna810lna，只需证1alnaa20，令ma1alnaa2，a0,4，····························································9分1a1则malna1lna，aa11因为ma0恒成立，所以ma在a0,4上单调递减，2aa1又m110，m2ln20，21由零点存在性定理得，a01,2使得ma00，即lna0，a0所以a0,a0时，ma0，ma单调递增，aa0,4时，ma0，ma单调递减，三校联考高三数学试题参考答案第6页共7页 11则mamaxma01a0lna0a021a0a02a03，···············11分aa001∵ya03在a01,2上显然单调递增，a0111∴a03230，a220∴ma0，即gx1gx210lna，得证．···············································12分三校联考高三数学试题参考答案第7页共7页